Suppose you have a 120-kg wooden crate resting on a wood floor. (a) what maximum force can you exert horizontally on the crate without moving it? (b) if you continue to exert this force once the crate starts to slip, what will the magnitude of its acceleration then be?

According the attached picture so we can assume that :

The coefficient of friction wood on wood are:

static friction μs = 0.5

dynamic friction μk = 0.2

a)The answer is:

F= 588.6 N

The explanation:

The maximum force  you can exert horizontally on the crate without moving it is equal:

Fsmax = μs * M * g

when μs is the static friction

and M i she mass

g is the gravitational force by substituted:

Fsmax = 0.5 * 120Kg * 9.81 m/s2

           = 588.6 N

when F= Fsmax

So the answer is F = 588.6 N

b)The answer is:

a= 2.943 m/s2

The explanation:

Fsmax is the force pushing the block , and there is a force of friction trying to stop the block .

So Fk = μk * N

when Fk = force of kinetic friction

μk = coefficient of kinetic friction of wood on wood

N= Normal force by substitution:

Fk= 0.2 * 120k *9.81 m/s2 = 235.44 N

so F = Fsmax - Fk

        588.6 - 235.44 = 353.16 N

when F = m*a

so 353.16 = 120Kg * a

so a ( the acceleration) = 353.16/120kg

                                      = 2.943 m/s2